The Pauli Exclusion Principle — Question 2

Question 2

An antisymmetric two-electron state has a normalization fixed by orthogonality

Let a and b be orthonormal one-electron spin orbitals. Consider $\Psi(1,2)=C[a(1)b(2)-b(1)a(2)]$, with C positive. The labels 1 and 2 include space and spin.

1. Normalize the state.

2. Apply particle exchange and determine what happens if b is replaced by a.

Question 2 - Solution

An antisymmetric two-electron state has a normalization fixed by orthogonality

(a) Method and calculation

$$\|\Psi\|^2=C^2[\langle a|a\rangle\langle b|b\rangle+\langle b|b\rangle\langle a|a\rangle-2|\langle a|b\rangle|^2]=2C^2.$$$$\boxed{C=1/\sqrt2}.$$

(b) Interpretation and further calculation

$$\boxed{\Psi(2,1)=-\Psi(1,2)}.$$$$b=a\Rightarrow\Psi=C[a(1)a(2)-a(1)a(2)]=0\Rightarrow\boxed{\text{duplicate occupation excluded}}.$$

Physical check

The zero function cannot be normalized. Exclusion concerns identical complete spin orbitals, so two electrons may share a spatial orbital if their spin states differ.